Math, asked by AshNegi, 11 months ago

plssssssssssssssssssssssss help

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Answered by neerajbhavya1784
1

sorry don't have enough time to solve all so I have solved only 1 question

12. (x-a) is the factor of x^{5} -a^{2}  x^{3} +2x+a+3

 then,   x-a=0       [ as factors have the remainder 0]

             x    =a

  p(x)= x^{5}-a^{2}  x^{3}+2x+a+3\\\\p(a)=a^{5}- a^{2} x^({3})+2a+a+3\\      = a^{5}-a^{5} +3a+3\\      = 0+3a+3\\       =3+3a\\

Now , a=3+3a

now lets factorise......

p(a)=x^{2} -2ax-3\\=x^{2}-2(3+3a) (x) -3\\=x^{2} -(6+6a)x-3\\=x^{2} -6x-6ax-3

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